InterviewSolution
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InterviewSolution
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`int_(0)^(2pi)|sinx|dx=?` |
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Answer» We know that `sinx` is positive, when `o le x le pi` and `sinx` is negative when `pi le x le 2pi`. `:.|sinx|={{:(sinx,"when" 0 le x le pi),(-sinx,"when" pi le x le 2pi):}` `:.int_(0)^(2pi)|sinx|dx=int_(0)^(pi)|sinx|dx+int_(pi)^(2pi)|sinx|dx` `=int_(0)^(pi)sinxdx+int_(pi)^(2pi)(-sinx)dx` `=[-cosx]_(0)^(pi)[cosx]_(pi)^(2pi)=(2+2)=4`. |
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